Question

Let +₆ (addition modulo 6) be a binary operation on S = {0, 1, 2, 3, 4, 5}. Write the value of $2{+}_6{4}^{-1}{+}_6{3}^{-1}.$

Answer 1

Here,

1 ${+}_6$ 1 = Remainder obtained by dividing 1 + 1 by 6
= 2

3 ${+}_6$ 4 = Remainder obtained by dividing 3 + 4 by 6
= 1

4 ${+}_6$ 5 = Remainder obtained by dividing 4 + 5 by 6
= 3

So, the composition table is as follows:

${+}_6$	0	1	2	3	4	5
0	0	1	2	3	4	5
1	1	2	3	4	5	0
2	2	3	4	5	0	1
3	3	4	5	0	1	2
4	4	5	0	1	2	3
5	5	0	1	2	3	4

We observe that the first row of the composition table coincides with the the top-most row and the first column coincides with the left-most column.
These two intersect at 0.
$\Rightarrow a{+}_{6}0=0{+}_{6}a=a,\forall a\in S$

So, 0 is the identity element.

From the table,
$$\begin{gathered} 4{+}_{6}2=0 \\ \Rightarrow 4^{-1}=2 \\ 3{+}_{6}3=0 \\ \Rightarrow 3^{-1}=3 \\ \text{Now,} \\ 2{+}_6{4}^{-1}{+}_6{3}^{-1}=2{+}_{6}2{+}_{6}3 \\ =4{+}_{6}3 \\ =1\mathit{} \end{gathered}$$